| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2009 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Sum/difference of two binomials simplification |
| Difficulty | Moderate -0.5 This is a straightforward multi-part question requiring routine application of binomial expansion for small n=4, followed by algebraic manipulation and basic calculus. Part (a) is direct computation, part (b) uses symmetry to cancel odd powers, and part (c) requires simple differentiation. All steps are standard textbook exercises with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| \((1+2x)^4 = 1 + 4(2x) + 6(2x)^2 + 4(2x)^3 + (2x)^4\) | M1 | (1), 4, 6, 4, (1) OE unsimplified with correct powers of \(x\). Algebraic multiplication must be a full method. |
| \(= 1 + 8x + 24x^2 + 32x^3 (+ 16x^4)\) | A1, A1, A1 | Accept \(a = 8\) provided 1st term is 1; \(b = 24\); \(c = 32\). 4 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| \((1 - 2y)^4 = 1 - 8x + 24x^2 - 32x^3 (+ 16x^4)\) | M1, A1ft | Replace \(x\) by \(-y\) even in M1 line of (a). PI ft c's non zero values for \(a\), \(b\) and \(c\) |
| \((1+2x)^4 + (1-2y)^4 = 1 + 8x + 24x^2 + 32x^3 + 16x^4 + 1 - 8x + 24x^2 - 32x^3 + 16x^4 = 2 + 48x^2 + 32x^4\) | A1cso | AG Be convinced. 3 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = 96x + 128x^3\) | M1 | A correct power of \(x\) OE |
| For st. pt. \(96x + 128x^3 = 0\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Since \(3 + 4x^2 > 0\) there is only one stationary point | E1 | Any valid explanation of curve having just one stationary point |
| The coordinates of the stationary point are \((0, 2)\) | B1 | \((0, 2)\) as the only stationary point. 4 marks total |
**5(a)**
$(1+2x)^4 = 1 + 4(2x) + 6(2x)^2 + 4(2x)^3 + (2x)^4$ | M1 | (1), 4, 6, 4, (1) OE unsimplified with correct powers of $x$. Algebraic multiplication must be a full method.
$= 1 + 8x + 24x^2 + 32x^3 (+ 16x^4)$ | A1, A1, A1 | Accept $a = 8$ provided 1st term is 1; $b = 24$; $c = 32$. 4 marks total
**5(b)**
$(1 - 2y)^4 = 1 - 8x + 24x^2 - 32x^3 (+ 16x^4)$ | M1, A1ft | Replace $x$ by $-y$ even in M1 line of (a). PI ft c's non zero values for $a$, $b$ and $c$
$(1+2x)^4 + (1-2y)^4 = 1 + 8x + 24x^2 + 32x^3 + 16x^4 + 1 - 8x + 24x^2 - 32x^3 + 16x^4 = 2 + 48x^2 + 32x^4$ | A1cso | AG Be convinced. 3 marks total
**5(c)**
$\frac{dy}{dx} = 96x + 128x^3$ | M1 | A correct power of $x$ OE
For st. pt. $96x + 128x^3 = 0$ | A1 |
$32x(3 + 4x^2) = 0$
Since $3 + 4x^2 > 0$ there is only one stationary point | E1 | Any valid explanation of curve having just one stationary point
The coordinates of the stationary point are $(0, 2)$ | B1 | $(0, 2)$ as the only stationary point. 4 marks total
**Total for Q5: 11 marks**
---5
\begin{enumerate}[label=(\alph*)]
\item By using the binomial expansion, or otherwise, express $( 1 + 2 x ) ^ { 4 }$ in the form
$$1 + a x + b x ^ { 2 } + c x ^ { 3 } + 16 x ^ { 4 }$$
where $a$, $b$ and $c$ are integers.
\item Hence show that $( 1 + 2 x ) ^ { 4 } + ( 1 - 2 x ) ^ { 4 } = 2 + 48 x ^ { 2 } + 32 x ^ { 4 }$.
\item Hence show that the curve with equation
$$y = ( 1 + 2 x ) ^ { 4 } + ( 1 - 2 x ) ^ { 4 }$$
has just one stationary point and state its coordinates.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2009 Q5 [11]}}